Introduction to Digital Signal and System Analysis - Tutorsindia

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Introduction to Digital Signal and System Analysis Z Domain Analysis 2 3 4 h[ n] = 0,1, α, α , α , α ... (5.8) If the impulse response decays with time then the system will return to its initial state before disturbance. Therefore the stability requires α < 1 in Eq.(5.8) as when n → ∞ makes h [ ∞] → 0 . i.e. in the z-plane, the real pole z = α must lie on the real axis and inside the unit circle. Consider a transfer function with conjugate imaginary poles: H ( z) = Y ( z) X ( z) = 1 ( z − jα )( z + jα ) (5.9) Rearranging, 2 2 z Y ( z) + α Y ( z) = X ( z) Or 2 Y ( z) + α z −2 Y ( z) = z −2 X ( z) Taking the inverse z-transform of both sides yields the difference equation 2 y[ n] + α y[ n − 2] = x[ n − 2] or 2 y[ n] = −α y[ n − 2] + x[ n − 2] (5.10) 2 From the above we use h[ n] = −α h[ n − 2] + d [ n − 2] to evaluate its impulse response, it can be obtained as 2 4 6 8 h[ n] = 0,0,1,0, −α ,0, α ,0, −α ,0, α ,... (5.11) It is clear that the stability condition also requires α < 1, in order to achieve that when n → ∞ , h [ ∞] → 0 . In the case of expression in polar co-ordinates, consider a conjugate pole pair, H ( z) = Y ( z) X ( z) = 1 2 2 ( z − r exp( jθ ))( z − r exp( − jθ )) z − 2rz cosθ + r = 1 (5.12) where r is the radius, q is the angle on the complex plane. The difference equation can be obtained in the same way as 72 Download free ebooks at bookboon.com
Introduction to Digital Signal and System Analysis Z Domain Analysis 2 y[ n] = 2r cosq y[ n −1] − r y[ n − 2] + x[ n − 2] . or 2 h[ n] = 2r cosq h[ n −1] − r h[ n − 2] + d[ n − 2] By evaluating the impulse response, assuming the system is causal, i.e. h [ n] = 0 when n < 0 h[0] = 0, h[1] = 0, h[3] = 2r cosq , h[2] = 1, h[4] = h[5] = 2 2 ( 2r cosq ) − r 2r cosq ( 2r cosq ) 2 2 2 2 2 2 ( − r ) − r 2r cosq = 2r cosq ( ( 2r cosq ) − r ) − r ) ... It can be found that the stable condition is he modulus |r|
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